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Finding Fourier Series Coefficients

Module by: Don Johnson. E-mail the author

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Summary: This module outlines the procedure for determining Fourier series coefficients.

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Assume for the moment that the Fourier series works. To find the Fourier coefficients, we note the following orthogonality properties of sinusoids.

Orthogonality

∀k,l:∫0Tsin2πktTcos2πltTdt=0 k l t 0 T 2 k t T 2 l t T 0 (1)
∫0Tsin2πktTsin2πltTdt=T2ifk=l∧k≠0∧l≠00ifk≠l∨k=0=l t 0 T 2 k t T 2 l t T T 2 k l k 0 l 0 0 k l k 0 l (2)
∫0Tcos2πktTcos2πltTdt=T2ifk=1∧k≠0∧l≠0Tifk=0=l0ifk≠l t 0 T 2 k t T 2 l t T T 2 k 1 k 0 l 0 T k 0 l 0 k l (3)
To use these, let's multiply the Fourier series for a square wave ( st= a 0 +∑k=1∞ a k cos2πktT+∑k=1∞ b k sin2πktT s t a 0 k 1 a k 2 k t T k 1 b k 2 k t T ) by cos2πltT 2 l t T and integrate. The idea is that, because integration is linear, the integration will sift out all but the term involving a l a l .
∫0Tstcos2πltTdt=∫0T a 0 cos2πltTdt+∑k=1∞ a k ∫0Tcos2πktTcos2πltTdt+∑k=1∞ b k ∫0Tsin2πktTcos2πltTdt t 0 T s t 2 l t T t 0 T a 0 2 l t T k 1 a k t 0 T 2 k t T 2 l t T k 1 b k t 0 T 2 k t T 2 l t T (4)
The first and third terms are zero; in the second, the only non-zero term in the sum results when the indices kk and ll are equal (but not zero), in which case we obtain a l T2 a l T 2 . If k=0=l k 0 l , we obtain a 0 T a 0 T . Consequently, ∀l,l≠0: a l =2T∫0Tstcos2πltTdt l l 0 a l 2 T t 0 T s t 2 l t T All of the Fourier coefficients can be found similarly.
a 0 =1T∫0Tstdt a 0 1 T t 0 T s t (5)
∀k,k≠0: a k =2T∫0Tstcos2πktTdt k k 0 a k 2 T t 0 T s t 2 k t T (6)
b k =2T∫0Tstsin2πktTdt b k 2 T t 0 T s t 2 k t T (7)

Exercise 1

The expression for a 0 a 0 is referred to as the average value of st s t . Why?

Solution

The average of a set of numbers is the sum divided by the number of terms. Viewing signal integration as the limit of a Riemann sum, the integral corresponds to the average.

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